Given:
kinetic energy of free particle, KE = 35ev
1eV = [tex]1.6\times 10^{-19}[/tex] J
mass of the particle, m = [tex]2\times 10^{-26}[/tex] Kg
accuracy in velocity= 0.2%= 0.002
Solution:
a) We know that
KE = [tex]\frac{1}{2}mv^{2}[/tex]
v = [tex]\sqrt{\frac{2KE}{m}}[/tex]
⇒ v = [tex]\sqrt{\frac{2\times 35\times1.6\times 10^{-19} }{2\times 10^{-26}}}[/tex]
v = [tex]2.36\times10^{4}[/tex] m/s
b) From Heisenberg's uncertainity principle:
[tex]\Delta x\Delta p = \frac{h}{4\pi}[/tex]
[tex]\Delta x.(mv) = \frac{h}{4\pi}[/tex]
[tex]\Delta x = \frac{h}{4\pi\times 2\times 10^{-26}\times2.36 \times 10^{4}\times 0.002}[/tex]
[tex]\Delta x = 0.56nm[/tex]
